Question

我知道在stackoverflow上多次询问此问题，但我无法找到可以解决问题的答案。

这是我从我的Activity的onPause（）函数调用的代码。

`

`protected void onPause() {
        // TODO Auto-generated method stub
        super.onPause();
        new DialogueTask().execute(this);   
        }

Dialogue类任务的实现是

class DialogueTask extends AsyncTask<Activity, Void, PopupWindow> {
    private View layout;

    @Override
    protected PopupWindow doInBackground(Activity... activities) {
        LayoutInflater layoutInflater = (LayoutInflater) activities[0]
                .getSystemService(Context.LAYOUT_INFLATER_SERVICE);

        layout = layoutInflater.inflate(R.layout.popup,
                (ViewGroup) activities[0].findViewById(R.id.popup_id));
        PopupWindow pw;
        pw = new PopupWindow(layout, 100, 100, true);
        return pw;
    }

    @Override
    protected void onPostExecute(PopupWindow pw) {
        // TODO Auto-generated method stub
        pw.showAtLocation(layout, Gravity.CENTER, 0, 0);
    }

这是popup.xml

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/popup_id"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:orientation="vertical" >

    <TextView
        android:layout_width="fill_parent"
        android:layout_height="wrap_content"
        android:layout_gravity="center_horizontal"
        android:text="Sure you want to Quit?" >
    </TextView>


</LinearLayout>

单击后退按钮，然后调用onPause（）并发出WindowLeaked异常。我想，我已经注意到通过实现AsyncTask在UI线程中显示了弹出窗口。那问题可能在哪里？

Answer 1

弹出窗口活动从主活动打开，您将退出主活动而不会忽略弹出窗口活动。 pw.showAtLocation（layout，Gravity.CENTER，0,0）之后的pw.dismiss（）将解决问题

@Override
protected void onPostExecute(PopupWindow pw) {
    // TODO Auto-generated method stub
    pw.showAtLocation(layout, Gravity.CENTER, 0, 0);
    pw.dismiss();
}

显示PopupWindow时出现WindowLeaked错误

1 个答案: